Question

A hammer of 50g moving at 50m/s strikes the nail in 0.01 seconds ,what is the force?

Answer 1

Given , mass of hammer

(m) = 500g = 0.5kg

Initial velocity of hammer

(u) = 50 m/s

Final velocity of hammer

(v) = 0 m/s

Duration (t) = 0.01 s

Using second law of motion ,

F = ma = m (v-u) / t

⇒ F = 0.5 ( 0 - 50 ) / 0.01

= (0.5 × - 50 ) / 0.01

= - 2,500 N

The - ve sign shows the force of 2500 N acts in the opposite direction of motion Given , mass of hammer

(m) = 500g = 0.5kg

Initial velocity of hammer

(u) = 50 m/s

Final velocity of hammer

(v) = 0 m/s

Duration (t) = 0.01 s

Using second law of motion ,

F = ma = m (v-u) / t

⇒ F = 0.5 ( 0 - 50 ) / 0.01

= (0.5 × - 50 ) / 0.01

= - 2,500 N

The - ve sign shows the force of 2500 N acts in the opposite direction of motion

Answer 2

By newton second law,

F=ma
50g is 0.5kg

since, a=v/t

F=m×v/t
= 0.5×50/0.01
=2500